Compactness of space of Lipschitz Continuous functionsCompactness of $(M, d)$ in terms of continuous functions $f:Mto mathbb R$Let $f$ be a continuous functions from $[0,1]$ to $mathbbR$. Then, $f$ is not necessarrily lipschitz.Compactness and Lipschitz functionsA strange criterion for compactnessCharacterizing compactness by properties of continuous functions defined on a spaceWhat is the relation between compactness , connectedness and continuous real-valued functions on $mathbbR^n$, n > 1?Does locally Lipschitz imply Lipschitz on closed balls?Are continuous mappings on a compact metric space Lipschitz?Closed & boundedness, sequentially compactness and CompletenessCompact subset of space of Continuous functions
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Compactness of space of Lipschitz Continuous functions
Compactness of $(M, d)$ in terms of continuous functions $f:Mto mathbb R$Let $f$ be a continuous functions from $[0,1]$ to $mathbbR$. Then, $f$ is not necessarrily lipschitz.Compactness and Lipschitz functionsA strange criterion for compactnessCharacterizing compactness by properties of continuous functions defined on a spaceWhat is the relation between compactness , connectedness and continuous real-valued functions on $mathbbR^n$, n > 1?Does locally Lipschitz imply Lipschitz on closed balls?Are continuous mappings on a compact metric space Lipschitz?Closed & boundedness, sequentially compactness and CompletenessCompact subset of space of Continuous functions
$begingroup$
Let $$X=f:[0,1]rightarrow[0,1], ftext is Lipschitz continuous $$with the supremum metric .
What can we say about the compactness of$ (X,d).$
I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.
I can't think of anything else. Kindly help !!
general-topology functional-analysis metric-spaces
$endgroup$
add a comment |
$begingroup$
Let $$X=f:[0,1]rightarrow[0,1], ftext is Lipschitz continuous $$with the supremum metric .
What can we say about the compactness of$ (X,d).$
I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.
I can't think of anything else. Kindly help !!
general-topology functional-analysis metric-spaces
$endgroup$
$begingroup$
"A space is compact iff every continuous real function is bounded." That is not true.
$endgroup$
– Mars Plastic
Mar 12 at 12:59
$begingroup$
Is there a different version of result ?@MarsPlastic
$endgroup$
– Devendra Singh Rana
Mar 12 at 13:04
$begingroup$
A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
$endgroup$
– Mars Plastic
Mar 12 at 13:06
add a comment |
$begingroup$
Let $$X=f:[0,1]rightarrow[0,1], ftext is Lipschitz continuous $$with the supremum metric .
What can we say about the compactness of$ (X,d).$
I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.
I can't think of anything else. Kindly help !!
general-topology functional-analysis metric-spaces
$endgroup$
Let $$X=f:[0,1]rightarrow[0,1], ftext is Lipschitz continuous $$with the supremum metric .
What can we say about the compactness of$ (X,d).$
I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.
I can't think of anything else. Kindly help !!
general-topology functional-analysis metric-spaces
general-topology functional-analysis metric-spaces
edited Mar 12 at 13:02
Devendra Singh Rana
asked Mar 12 at 10:56
Devendra Singh RanaDevendra Singh Rana
7871416
7871416
$begingroup$
"A space is compact iff every continuous real function is bounded." That is not true.
$endgroup$
– Mars Plastic
Mar 12 at 12:59
$begingroup$
Is there a different version of result ?@MarsPlastic
$endgroup$
– Devendra Singh Rana
Mar 12 at 13:04
$begingroup$
A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
$endgroup$
– Mars Plastic
Mar 12 at 13:06
add a comment |
$begingroup$
"A space is compact iff every continuous real function is bounded." That is not true.
$endgroup$
– Mars Plastic
Mar 12 at 12:59
$begingroup$
Is there a different version of result ?@MarsPlastic
$endgroup$
– Devendra Singh Rana
Mar 12 at 13:04
$begingroup$
A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
$endgroup$
– Mars Plastic
Mar 12 at 13:06
$begingroup$
"A space is compact iff every continuous real function is bounded." That is not true.
$endgroup$
– Mars Plastic
Mar 12 at 12:59
$begingroup$
"A space is compact iff every continuous real function is bounded." That is not true.
$endgroup$
– Mars Plastic
Mar 12 at 12:59
$begingroup$
Is there a different version of result ?@MarsPlastic
$endgroup$
– Devendra Singh Rana
Mar 12 at 13:04
$begingroup$
Is there a different version of result ?@MarsPlastic
$endgroup$
– Devendra Singh Rana
Mar 12 at 13:04
$begingroup$
A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
$endgroup$
– Mars Plastic
Mar 12 at 13:06
$begingroup$
A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
$endgroup$
– Mars Plastic
Mar 12 at 13:06
add a comment |
1 Answer
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$begingroup$
If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.
$endgroup$
add a comment |
$begingroup$
If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.
$endgroup$
add a comment |
$begingroup$
If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.
$endgroup$
If $f_n(x)=x^n$ ($xin[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_ninmathbb N$ has no convergente subsequence. Therefore, your space is not compact.
answered Mar 12 at 11:02
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
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$begingroup$
"A space is compact iff every continuous real function is bounded." That is not true.
$endgroup$
– Mars Plastic
Mar 12 at 12:59
$begingroup$
Is there a different version of result ?@MarsPlastic
$endgroup$
– Devendra Singh Rana
Mar 12 at 13:04
$begingroup$
A topological space $X$ with the property that every continuous $fcolon Xto Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview.
$endgroup$
– Mars Plastic
Mar 12 at 13:06